It seems to make intuitive sense that if the events $A$ and $B$ are independent then $P(A|B\cap C)=P(A|C)$ because the occurrence of event $B$ should not change the probability of event $A$ even when we know of the occurrence of another event $C$, however I can't seem to prove or disprove the above claim so I'm prompted to ask here. If the answer is yes then please provide a proof, if the answer is no then please provide a counterexample.


2 Answers 2


This conjecture is not true. One easy way to see this is that while $P(A|C)$ is well defined as long as $P(C) > 0$, $P(A|B\cap C)$ could well be undefined because $P(B \cap C)$ may vanish (e.g., when $B$ and $C$ are disjoint), regardless whether $A$ and $B$ are independent.

To give a non-trivial example when both sides are well defined, let $X, Y \text{ i.i.d. } \sim N(0, 1)$, then define $A = \{X > 0\}, B = \{Y > 0\}$ and $C = \{X + Y > 0\}$. Because $X$ and $Y$ are independent, clearly $A$ and $B$ are independent. Since \begin{align*} & P(A \cap C) = P(X > 0, X + Y > 0) = \int_0^\infty P(x + Y > 0)\varphi(x)dx \\ =& \int_0^\infty \Phi(x)\varphi(x)dx = \int_{1/2}^1udu = \frac{3}{8}, \end{align*} and $P(C) = \frac{1}{2}$, we have \begin{align*} & P(A | C) = \frac{P(A \cap C)}{P(C)} = \frac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4}, \\ & P(A | B \cap C) = \frac{P(X > 0, Y > 0, X + Y > 0)}{\frac{3}{8}} = \frac{P(X > 0, Y > 0)}{\frac{3}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}} = \frac{2}{3}, \end{align*} disproves your conjecture.

Another easier-to-compute discrete counterexample: let $X, Y \text{ i.i.d. } \sim B(1, 0.5)$, then define $A = \{X = 1\}, B = \{Y = 1\}, C = \{X + Y = 1\}$. It is then easy to verify $P(A | C) = \frac{1}{2}$ while $P(A | B \cap C) = 0$.

In fact, when all probabilities under consideration are positive, the equality $P(A | C) = P(A | B \cap C)$ is equivalent to $P(A \cap B | C) = P(A | C) P(B | C)$, which is the definition of $A$ and $B$ are conditionally independent given $C$. I am sure you understand that conditional independence and unconditional independence are two different things, and you could find numerous counterexamples that one does not imply the other in this forum.

  • $\begingroup$ I'm sorry, I'm new to the concept of conditional independence, I'm working through an elementary probability theory textbook as I asked this question. That being said, I seem to have neglected some assumptions, such as the requirement that the probabilities of all events involved are indeed positive. But if we also assume that all the events are non-disjoint, that is $A \cap B \cap C \neq \emptyset$ would that change anything? If not then under what assumptions can unconditional independence imply conditional independence? $\endgroup$
    – P.K.
    Nov 10, 2023 at 13:29
  • 1
    $\begingroup$ @P.K. The first normal counterexample satisfies $A \cap B \cap C \neq \varnothing$, and it shows you that your conjecture is still not true. There is no general condition to make these two concepts coincide, because they are just different concepts (but you might check the corollary in this answer). $\endgroup$
    – Zhanxiong
    Nov 10, 2023 at 13:31
  • $\begingroup$ In terms of events, a trivial sufficient condition for unconditional independence implying conditional independence is $A, B, C$ are jointly independent. But this condition tends to be too strong. $\endgroup$
    – Zhanxiong
    Nov 10, 2023 at 13:39
  • $\begingroup$ I'm going to present the specific problem that I encountered for clarity. Consider an urn containing $N$ balls with different colors. The number of red balls, $X$, has a binomial distribution $X \sim B(N,p)$ with $p$ known. We draw a random ball from the urn, note its color, then return it and draw a second ball. Let $F_1$ be the event that the first ball is red and $F_2$ be the event that the second ball is red. It is clear that $F_1$ and $F_2$ are independent. It is then claimed that $P(F_2 | X=n \cap F_1) = P(F_2 | X=n)$. Why is this true here but not in general? $\endgroup$
    – P.K.
    Nov 10, 2023 at 13:57
  • $\begingroup$ And to make it clear, the events $A, B, C$ In the question above correspond to the events $F_2, F_1, \{X=n\}$ respectively. $\endgroup$
    – P.K.
    Nov 10, 2023 at 13:59

No it is not true in general.

Indeed it is possible that

$B \cap C = \emptyset$ and

$A \cap C >0$,

so that $P(A|B \cap C) \neq P(A|C)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.